2008/01/24

Pure Mathematics 補底

1.
If f(x) is a polynomial of degree 2, and f(1) = f(2) = 0, then it is NOT true to say
f(x) = (x-1)(x-2).
Instead, we should write
f(x) k(x-1)(x-2) for some non-zero constant k.
[Read : 2004-AL-pure math-paper I-Q.4]

2.
Let f(x) be a polynomial of degree not less than 3.
If f(x) is divided by (x-1)(x-2)(x-3), then it is NOT true to say
f(x) = (x-1)(x-2)(x-3)Q(x) + R
(where R is a constant, the remainder)
Instead, we should write
f(x) (x-1)(x-2)(x-3)Q(x) +
that is,
we should set the remainder as a polynomial of degree 2
(just less than the degree of the divisor (x-1)(x-2)(x-3) by 1)

4.
If , it is NOT true to draw the following conclusion.
But, if we know , then it is true to have .
Proof: Multiply and , we have and result follows.
[Read : 1999-AL-pure math-paper I-Q.2]

5.
Let {} be a sequence of real numbers.
Knowing that .
It is NOT true to say
{} is increasing.
Just give an example.
Obviously, (for n = 1,2,3,4,5), but {} is not increasing.
[Read 1999-AL-pure math-paper I-Q.2]

6.
Let {} be a sequence of real numbers.
Knowing that .
We cannot say {} is bounded from above by . is NOT an upper bound of the sequence {}.
Instead, with a small effort added, we have for all positive integers n.
And now, it is clear that, 1 is an upper bound of {}.

7.
(a) Solve
Solution:
(b) Solve
Solution: or

8.
Consider the quadratic equation
Given that the discriminant .
Then it is NOT true to say that the quadratic equation has no real solution.
Just consider a quadratic equation with roots 0 and i.
That is
Then .
But there is a real root (x = 0) for the quadratic equation.

9.
“Vector" is out of the syllabus now.
“Linearly independent", “linearly dependent" are out-of-syllabus concepts.

11.
Symbols about left and right hand limits
“" does NOT mean “"
“" does NOT mean “"
See the difference? Try to obtain the following by sight.
(a)
(b) does not exist.
(c) does not exist (since the left and right lmits are not the same)
(d)

12.
To evaluate , please do NOT split it as
No! Both limits may NOT exist!
You may try “rationalization", “sum to product", “MVT" etc.

13.
How to evaluate = ?
Try the Mean Value Theorem (of course, should be differentiable at certain interval), that is
= for some
= for some
= for some
(if is bounded, then the limit is zero, read 10.(e))

15.
Be careful, the following may NOT be true!
e.g. Given , evaluate
It is WRONG to write
= (for )
= 1
You should give somthing like
=
=
= (l’hôpital rule)
= 0 (see, NOT 1)
[Read : 2002-AL-pure math-paper II-Q.3]

16.
Please please please DON’T make the following silly mistakes!
(a)
(b)
For (a), try the reduction formula.
For (b), there is no closed form, just stop there.

17.
The following definitions MUST be memorized!
(a) “ is well-defined at " means “ can be found".
(b) “ is continuous at " means ““.
(c) “ is differentiable at " means “ exists".

18.
Q: At a look at the definitions above, why there is no mention of the left and right hand limits? When should we consider the left and right hand limits?
A: When has DIFFERENT expressions for and (say)
e.g.
(a) Let for and for . Then is continuous at . Proof: while Hence is continuous at .
(b) Let for and for with f(3) = 5 Then is NOT continuous at . (why?)
(c) Let . To prove is continuous at , it is NOT necessary to “divide cases", since both left and right hand limits are zero. i.e. , hence and therefore, . Done.

19.
Given

for for

(both and are ‘nice’ functions)
Then, to find , the problem only appears at , hence DON’T write

23.
Many questions involve the Mean Value Theorem in Cauchy form, that is
both and are continuous on [a , b] and differentiable on (a , b) with , then
then for some (a , b).
DON’T write the following as a so-called proof

for some (a , b)

NO! We CANNOT ensure both in and are the same!!! That is, all we have may be the following

for some (a , b) for some (a , b)

and then we can do nothing at all…How to prove? Try the read
[Read 2006-AL-pure math-paper II-Q.11(a)]

Thank you for the invitation. However, it may be too much for me to maintain my blog and forum, not to mention to be a person-in-charge in one of the sections in your forum. Instead, I ‘d registered as one of your members and I’ll answer questions or just share something in your forum ocassionally.